Answer:
[tex] x^2 +(y-3)^2 = 25[/tex]
And we want to find the coordinate axes so then we can do the following:
If x=0 we have:
[tex] 0^2 +(y-3)^2 = 25[/tex]
[tex] (y-3)^2= 25[/tex]
[tex] y-3= \pm 5[/tex]
[tex] y_1 = 5+3=8[/tex]
[tex] y_2 = -5+3=-2[/tex]
Now of y =0 we have:
[tex] x^2 +9 = 25[/tex]
[tex] x^2 = 16[/tex]
[tex] x= \pm 4[/tex]
And then the coordinate axes are:
(4,0) (-4,0), (0,8), (0,-2)
Step-by-step explanation:
For this cae we have the following functon given:
[tex] x^2 +(y-3)^2 = 25[/tex]
And we want to find the coordinate axes so then we can do the following:
If x=0 we have:
[tex] 0^2 +(y-3)^2 = 25[/tex]
[tex] (y-3)^2= 25[/tex]
[tex] y-3= \pm 5[/tex]
[tex] y_1 = 5+3=8[/tex]
[tex] y_2 = -5+3=-2[/tex]
Now of y =0 we have:
[tex] x^2 +9 = 25[/tex]
[tex] x^2 = 16[/tex]
[tex] x= \pm 4[/tex]
And then the coordinate axes are:
(4,0) (-4,0), (0,8), (0,-2)
